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klemol [59]
2 years ago
11

Which of the following is a true proportion of the figure based on the triangle

Mathematics
1 answer:
Anni [7]2 years ago
4 0

Based on the triangle proportionality theorem, the true proportion is h/j=k/i

<h3>How to solve for the true proportion</h3>

In the diagram that we have here, we have TU to be parallel to QR.

We have following

ST is equal to h, TQ is equal to j, SU is equal to k and UR is equal to i.

This can then be used to form the equation

ST/TQ = SU/UR

When we h/j = k/i

Read more on proportionality here:

brainly.com/question/20384391

#SPJ1

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Will mark Brainliest! You MUST explain step-by-step how you solved the problem.
sergiy2304 [10]

Answer:

choice C) 30 ft

Step-by-step explanation:

tan 41° = height/34

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3 0
3 years ago
If h(-8), find 1/3*h(x) = x^2-5x+7
Lynna [10]

The value is 333

<h3>How to determine the function</h3>

From the information given, we have:

  • x = - 8
  • 1/3*h(x) = x^2-5x+7

Now, let's substitute the value of 'x' in the function:

1/3*h(x) = x^2-5x+7

1/ 3 × h(-8) = ( - 8)² - 5 ( -8) + 7

Make 'h ( -8)' the subject of formula

h ( -8) = \frac{( -8)^2 - 5(-8) + 7}{1/ 3}

h ( -8) = \frac{64 + 40+ 7 }{1/ 3}

Take the sum of the numerator

h ( -8) = \frac{111}{1/3}

Take the inverse of the denominator and multiply

h ( -8) = 111 × 3/ 1

h ( -8) = 333

We can see that through the substitute of the value of x as - 8, we get 333

Thus, the value is 333

Learn more about algebraic expressions here:

brainly.com/question/723406

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7 0
2 years ago
solve the systems of equations via graphing then check the solution in each equation: 4x+6y=18, (y+2)=1/3(x+2)
Mariulka [41]

just wondering are you still stuck

8 0
3 years ago
a jogger run 3 miles north and 5.5 miles east . what is the shortest distance back to their starting point
r-ruslan [8.4K]
You subtract 3 miles from the 5.5 miles which gives you  2.5

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Solve for y x=y^2+4y
Lorico [155]
x=y^2+4y\\ y^2+4y-x=0\\\Delta=4^2-4\cdot1\cdot(-x)=16+4x\\\\&#10;1.\ \Delta0\\&#10;\sqrt{\Delta}=\sqrt{16+4x}=\sqrt{4(4+x)}=2\sqrt{x+4}\\&#10;y_1=\frac{-4-2\sqrt{x+4}}{2\cdot1}=-2-\sqrt{x+4}\\&#10;y_2=\frac{-4+2\sqrt{x+4}}{2\cdot1}=-2+\sqrt{x+4}\\

-----------------------------------------------------

x=y^2+4y\\ y^2+4y-x=0\\&#10;y^2+4y+4-4-x=0\\&#10;(y+2)^2=x+4\\&#10;y+2=\sqrt{x+4} \vee y+2=-\sqrt{x+4}\\&#10;y=-2+\sqrt{x+4} \vee y=-2-\sqrt{x+4}\\

4 0
4 years ago
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